The question: Consider the surface given by $x^2 +y^2 =z^2$ where $0\leq z\leq1$
(a) Parametrise the surface.
(b) Use surface integrals of the first kind to compute its area. Explain your work.
So far solution was
(a) $x=z\cos \theta, y=z\sin \theta, z=z$
$r^2\cos ^2\theta + r^2\sin ^2\theta =z^2$
$r^2=z^2$ so $z=r$ with $0\leq r\leq1$
(b) $S=\int^{2\pi}_{0} \int^{1}_{0}rdrd\theta $
But I don't think that's the correct integral, any help would be greatly appreciated thank you
Start with the standard expression below for the surface integral,
$$ I=\int_S \sqrt{1+(z_x^{'})^2+(z_y^{'})^2 }dxdy$$
where, use the given surface parametrization $x^2+y^2=z^2$ directly, the integrand is
$$ \sqrt{1+(z_x^{'})^2+(z_y^{'})^2 }= \sqrt{1+(-x/z)^2+(-y/z)^2}=\sqrt{2}$$
Because of the circular boundary in the $xy$-plane, convert to the polar coordinates for the integration,
$$ I=\int_S \sqrt{2}dxdy=\sqrt{2}\int_0^{2\pi}\int_0^1 rdrd\theta=\sqrt{2}\pi$$